Calculus I: MATH 1351-030
Fall 2011
Exam 3 (C)
Name:
Instructions. (Part 1) Solve the following problem. Show your work clearly. You must
write out all relevant steps. Simply having the correct answer does not give you
credit.
1. A ball is dropped (initial velocity v0 = 0) from the top of a building and falls for 2
seconds before hitting the ground.
(a.) What is the height of the building?
(b.) What is the velocity of the ball at the moment it hits the ground?
Solution:
(a.) The function for position of a falling body at time t is s(t) = 12 gt2 + v0 t + s0 , where
g = −32f t/s2 . In this case we have v0 = 0, so s(t) = 21 gt2 + s0 . The ball hits the
ground when s(t) = 0. Since it hits in 2 seconds, we have
1
g(2)2 + s0 = 0
2
1
0 = − (32)(2)2 + s0
2
1
(32)(2)2
s0 =
2
s0 = 64 f t
s(2) =
(b.) From part (a), we have the position function s(t) = 21 gt2 + 64 = −16t2 + 64.
Differentiating, we get the velocity function to be v(t) = −32t. So at time t = 2
when the ball hits the ground, the velocity is v(2) = −64 f t/s.
Calculus I: MATH 1351-030/Exam 3 (C) – Page 2 of 9 – Name:
2. Use logarithmic differentiation to find
dy
dx
for
(x2 + 1) (x + 3)1/2
.
y=
x−1
(Your answer should be in terms of only x0 s.)
Solution:
ln y =
=
=
1 dy
=
y dx
dy
=
dx
dy
=
dx
!
(x2 + 1) (x + 3)1/2
ln
x−1
ln x2 + 1 + ln(x + 3)1/2 − ln(x − 1)
1
ln x2 + 1 + ln(x + 3) − ln(x − 1)
2
2x
1
1
+
−
2
x + 1 2(x + 3) x − 1
2x
1
1
y·
+
−
x2 + 1 2x + 6 x − 1
(x2 + 1) (x + 3)1/2
2x
1
1
+
−
x−1
x2 + 1 2x + 6 x − 1
Calculus I: MATH 1351-030/Exam 3 (C) – Page 3 of 9 – Name:
3.
(a) State the limit definition of the derivative.
(b) Using the limit definition of the derivative, find the derivative of f (x) = x2 − 2x.
Solution:
(a) For a function f (x), the derivative is given by
f (x + ∆x) − f (x)
∆x→0
∆x
f 0 (x) = lim
if the limit exists.
(b) The derivative of f (x) = x2 − 2x is
f 0 (x) =
=
=
=
=
=
f (x + ∆x) − f (x)
∆x→0
∆x
[(x + ∆x)2 − 2(x + ∆x)] − (x2 − 2x)
lim
∆x→0
∆x
x2 + 2x∆x + ∆x2 + −2x − 2∆x − x2 + 2x
lim
∆x→0
∆x
2
2x∆x − 2∆x + ∆x
lim
∆x→0
∆x
∆x (2x − 2 + ∆x)
lim
∆x→0
∆x
lim 2x − 2 + ∆x
lim
∆x→0
= 2x − 2
Calculus I: MATH 1351-030/Exam 3 (C) – Page 4 of 9 – Name:
4. (bonus) Show using implicit differentiation that
d
1
sin−1 (x) = √
.
dx
1 − x2
Solution: (First, note that I fixed the typo that I pointed out during the exam.) Let
y = sin−1 x. Then:
y = sin−1 x
sin y = x
d
d
sin y =
x
dx
dx
dy
cos y
= 1
(implicit differentiation)
dx
dy
1
=
dx
cos y
We can√see from drawing a reference triangle that if sin y = x, then we must have
cos y = 1 − x2 .
Therefore, we have
dy
1
= √
dx
1 − x2
Calculus I: MATH 1351-030/Exam 3 (C) – Page 5 of 9 – Name:
Instructions. Part 2 Solve each of the following problems. Choose the best solution to each
problem and clearly mark your choice.
1. Find the derivative of the following function:
f (x) = sin (x2 + x) + e3x
(a) f 0 (x) =
cos (x2 +x)
2x+1
− 3e3x
(b) f 0 (x) = 0
(c) f 0 (x) = cos (x2 + x) + e3x
(d) f 0 (x) = − cos (x2 + x) + e3x
(e)
4 f 0 (x) = (2x + 1) cos (x2 + x) + 3e3x
dy
for y =
dx
dy
= 6x
dx
x
dy
= ln6 6
dx
dy
= 6x ln 6
dx
2. Find
(a)
(c)
(e)
4
6x .
(b) Derivative does not exist.
dy
(d) dx
= 6x log6 6
3. If y = x3 + 6x2 − 5x + 10, what is
(a) x
(c)
4 0
(e) 6
d4 y
?
dx4
(b) The 4th derivative does not exist.
(d) x3
4. Find the derivative of the following function:
g(x) = x2 ln x − sec x
(a)
(b)
(c)
(d)
4
(e)
dg
dx
dg
dx
dg
dx
dg
dx
dg
dx
=
1
x
− 2x + sec2 x
= x − 2x ln x + sec x tan x
= x + 2x ln x − tan2 x
= x + 2x ln x − sec x tan x
= x − 2x ln x + tan x
Calculus I: MATH 1351-030/Exam 3 (C) – Page 6 of 9 – Name:
dy
for y = xx .
dx
dy
= (ln x + 1)xx
dx
5. Find
(a)
4
(c) 1
dy
(e) dx
=
(b)
(d)
1
xx
6. If y = ex (x − 1), then y 00 (0) equals
(a) -2
(c) 0
(e)
4 1
dy
dx
dy
dx
= xx
= ln x + 1
(b) -1
(d) none of these
7. Find the derivative of f (x) = cos−1 (4x + 3).
−4
4
(a) f 0 (x) = 1+(4x+3)
(b) f 0 (x) = 1+(4x+3)
2
2
−1
−4
0
0
√
√
(d) f (x) =
(c)
4 f (x) =
2
0
(e) f (x) =
1−(4x+3)2
1−(4x+3)
√ 1
1+(4x+3)2
8. Find the derivative of the following function:
f (x) =
(a) f 0 (x) = x − 8
3
(c) f 0 (x) = −x x−x+8
3
−3x2 −6x+1
0
(e) f (x) =
2x
x3 + 3x2 − x + 4
x2
(b)
4 f 0 (x) =
(d) f 0 (x) =
x3 +x−8
x3
3x2 +6x−1
2x
9. The position of an object at time t is given by the function
s(t) = t3 − 9t2 + 24t − 6.
Over the time interval [0, 8], when is the object accelerating and when is it decelerating?
(a) Accelerating on [0, 3) ∪ (3, 8]; decelerating at t = 3.
(b)
4 Accelerating on (3, 8]; decelerating on [0, 3).
(c) Accelerating on [0, 3); decelerating on (3, 8].
(d) Accelerating on [0, 2) ∪ (4, 8]; decelerating on (2, 4).
(e) Accelerating on [0, 8]; never decelerating.
10. Find an equation for the tangent line at the prescribed point for the following function:
f (x) = ex cos x
(a) y = 1
(c)
4 y =x+1
(e) y = x + π
at x = 0
(b) y = 2x
(d) y = x
Calculus I: MATH 1351-030/Exam 3 (C) – Page 7 of 9 – Name:
11. Find the derivative of g(x) = log x.
(a)
4 g 0 (x) = x ln1 10
(c) Derivative does not exist
(e) g 0 (x) = x1
(b) g 0 (x) =
(d) g 0 (x) =
12. Let y be a differentiable function of x. Find
(a)
(c)
4
(e)
dy
dx
dy
dx
dy
dx
2
= 3xy2−1
2
= xy2
= 3x2 − 3y 2
dy
dx
(b)
(d)
ln 10
x
1
ln x
if x3 − y 3 = 1.
dy
dx
dy
dx
= 3x2
√
3
= 3x2
13. The position of an object at time t is given by the function s(t) = 3t2 − 2t + 2. What is
the velocity of the object at time t?
(a)
4 v(t) = 6t − 2
(b) v(t) = |6t − 2|
(c) v(t) = −6t + 2
(d) v(t) = 6
(e) v(t) = 3t − 2
14. Find the derivative of the following function:
h(x) = tan x − csc x + tan−1 x
1
1+x2
(a)
dh
dx
= cot2 x − csc x cot x +
(b)
dh
dx
= sec x + csc x cot x +
(c)
dh
dx
= sec2 x − csc x cot x +
1
1−x2
(d)
dh
dx
= csc2 x + sec x tan x +
√ 1
x2 −1
(e)
4
dh
dx
= sec2 x + csc x cot x +
1
1+x2
1
1−x2
15. The position of an object at time t is given by the function s(t) = sin(t). What is
the average velocity over the time interval 0 ≤ t ≤ π/2 and what is the instantaneous
velocity at time t = π/4?
(a) Average velocity is −2/π; instantaneous velocity is v(π/4) = 0.
√
(b) Average velocity is 1/π; instantaneous velocity is v(π/4) = − 2/2.
√
(c) Average velocity is 0; instantaneous velocity is v(π/4) = − 2/2.
√
(d)
4 Average velocity is 2/π; instantaneous velocity is v(π/4) = 2/2.
√
(e) Average velocity is −2/π; instantaneous velocity is v(π/4) = 2/2.
Calculus I: MATH 1351-030/Exam 3 (C) – Page 8 of 9 – Name:
16. Which of the following best describes when a function f (x) is differentiable at a point
x = c?
(x)
(a) The following limit exists for x = c: lim∆x→0 f (∆x)−f
x
(b) f (x) is defined at x = c.
(c) f (c) 6= 0.
(d) f (x) is continuous at x = c.
(e)
4 The following limit exists for x = c: lim∆x→0
3
1
.
17. Find the derivative of g(t) = 1−2t
(a)
4 g 0 (t) =
(c) g 0 (t) =
(e) 0
f (x+∆x)−f (x)
∆x
(b) g 0 (t) =
(d) g 0 (t) =
6
(1−2t)4
1
−6(1−2t)3
−3
(1−2t)4
3
(1−2t)4
√
18. Find y 0 for y = 2x4 − x3 + x − 9
(a)
4 y 0 = 8x3 − 3x2 + 2√1 x
√
(b) y 0 = 8x3 − 3x2 + 3 x − 1
(c) y 0 = x3 − x2 −
1
√
2 x
(d) y 0 = 8x3 − 3x2 +
1√
x
2
√
3
(e) y 0 = 8x3 + 3x2 + 2 x2
19. The function f (x) whose graph is shown has f 0 (x) = 0 at which values of x?
(a) x = a, b and d
(c)
4 x = a only
(e) x = a and c
(b) x = b and d
(d) x = a, b, c, and d
Calculus I: MATH 1351-030/Exam 3 (C) – Page 9 of 9 – Name:
20. Which of the following are true statements?
I. If f (x) is continuous, then it is differentiable.
II. If f (x) is differentiable, then it is continuous.
III. If f (x) is defined for all real x, then it is differentiable.
(a) I
(c)
4 II
(e) I, II, and III
(b) I and II
(d) III